chapter9 - Chapter 9 Solutions to 9.1 1. F(s) = e st e 2t...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 9 Solutions to 9.1 1. F(s) = e st e 2t dt = lim n n e (2s)t dt = 1 2 s lim n [ ] e (2s)t n = 1 s 2 if s > 2. 3. F(s) = e st sin btdt = lim n n e st sin btdt = lim n e st [sin bt bcos bt] s 2 + b 2 n = b s 2 + b 2 if s > 0. 5. F(s) = e st cosh btdt = lim n n e st e bt + e bt 2 dt = lim n e (b s)t 2(b s) e (b + s)t 2(b + s) n = lim n e (b s)n 1 2(b s) e (b + s)n + 1 2(b + s) = s s 2 b 2 where s > |b|. 7. F(s) = e st (2t)dt = 2lim n n e st tdt = 2lim n e st (st 1) s 2 n = 2 lim n n se sn 1 s 2 e sn + 1 s 2 = 2 s 2 where s > 0. 9. F(s) = 2 e st (1)dt + 2 e st (1)dt = e st s 2 + lim n 2 n (e st )dt = [ e 2s s + 1 s ] + lim n e st s 2 n = 1 s 1 e 2s s + lim n [ 1 se sn 1 se 2s ] = 1 s (1 2e 2s ) provided s > 0. 11. F(s) = e st (e t sin t)dt = lim n n e (1 s)t sin tdt = lim n e (1 s)t [sin t cos t] (s 1) 2 + 1 n = lim n e (1 2)n [sin n cos n] (s 1) 2 + 1 + 1 (s 1) 2 + 1 = 1 (s 1) 2 + 1 where s > 1. 267 268 Ch. 9 The Laplace Transform and Some Elementary Applications 13. L[2t e 3t ] = L[2t] L[e 3t ] = 2L[t] 1 s 3 = 2( 1 s 2 ) 1 s 3 = 2 s 2 1 s 3 , where s > 3. 15. L[cosh bt] = L e bt 2 + e bt 2 = 1 2 L[e bt ] + 1 2 L[e bt ] = 1 2 1 s b + 1 2 1 s + b = s s 2 b 2 where s > |b|. 17. L[3t 2 5cos 2t + sin 3t] = 3L[t 2 ] 5L[cos 2t] + L[sin 3t] = 3 2! s 3 5 s s 2 + 4 + 3 s 2 + 9 = 6 s 3 5s s 2 + 4 + 3 s 2 + 9 . 19. L[2e 3t + 4e t 5sin t] = 2L[e 3t ] + 4L[e t ] 5L[sin t] = 2 s + 3 + 4 s 1 5 s 2 + 1 where s > 1. 21. L[4cos 2 bt] = 2L[2cos 2 bt] = 2L[cos 2bt + 1] = 2L[cos 2bt] + 2L[1] = 2 ( s 4b 2 + s 2 ) + 2 ( 1 s ) = 4(s 2 + 2b 2 ) (s 2 + 4b 2 )s where s > 0. 23. The graph of f is piecewise continuous on [0, ). 25. The graph of f is not piecewise continuous on [0, ), because lim n 1 + f(t) = , i.e. the right hand limit as t approaches 1 from above is not finite. 27. The graph of f is piecewise continuous on [0, )....
View Full Document

Page1 / 26

chapter9 - Chapter 9 Solutions to 9.1 1. F(s) = e st e 2t...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online